Two calculus students attempt to “break” math.

Work in groups of 3–4, writing your answers on a separate sheet of paper.

Check out this dialogue between two calculus students (based on a true story):

Devyn
Riley, I have something very important to say.
Riley
Yeah? Hit me with it.
Devyn
I think I just broke math.
Riley
I’ve suspected for ages that all this calculus stuff was razzmatazz. Lay it on me.
Devyn
Consider the vector field:
\[ \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}} \]
Riley
Got it. It looks like a whirlpool.
Devyn
I know! Now compute its curl.
Riley
OK–I get zero curl.
Devyn
I know! Now consider Green’s Theorem.
Riley
You mean:
\[ \iint _R \curl \vec {F} \d A = \oint _{C} \vec {F}\dotp \d \vec {p} \]
What’s \(R\)? What’s \(C\)?
Devyn
Let \(R\) be the unit disk centered at the origin.
Riley
OK, so \(C\) is the unit circle centered at the origin.
Devyn
Right. Now here’s the deal…The left-hand side of the equation is zero because the curl of our vector field is zero.
Riley
Oh. And the right-hand side of the equation cannot be zero because our vector field looks like a whirlpool.
Devyn
And now we have zero equals something not zero, and kablamy, math is in ruins.
What just happened? Explain why Devin and Riley think math is broken. Try to take it step-by-step.

Let’s investigate further. To really understand this, we’re going to have to check each of their claims.

Consider the vector field
\[ \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}} \]
Plot some vectors on the grid below. Focus on getting the directions of the vectors correct, and don’t worry to much about the magnitudes. Does it look like a “whirlpool?”
Letting \(\vec {F}(x,y) = \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}}\), explain why someone might believe that
\[ \oint _{C} \vec {F}\dotp \d \vec {p} \]
is nonzero. No computation are necessary at this point.
Parametrize the unit circle \(U\) centered at the origin and compute:
\[ \oint _{U} \vec {F}\dotp \d \vec {p} \]
Parametrize a circle \(C\) of radius \(r\) centered at the origin and compute:
\[ \oint _{C} \vec {F}\dotp \d \vec {p} \]
As a gesture of friendship, we reveal that:
\begin{align*} \grad \arctan (y/x) &= \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}}\\ \grad \arctan (-x/y) &= \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}} \end{align*}

Confirm these equations.

Use the Fundamental Theorem of Line Integrals to compute
\[ \oint _S \frac {-y}{x^2+y^2}\d x + \frac {x}{x^2+y^2}\d y \]
where \(S\) is the square with vertices \((1,1)\), \((-1,1)\), \((-1,-1)\), and \((1,-1)\) drawn in a counterclockwise fashion.
For the fundamental theorem to apply, the chosen path must be in the domain of the potential function.
Use the Fundamental Theorem of Line Integrals to compute
\[ \oint _Q \frac {-y}{x^2+y^2}\d x + \frac {x}{x^2+y^2}\d y \]
where \(Q\) is the square with vertices \((a,a)\), \((-a,a)\), \((-a,-a)\), and \((a,-a)\) drawn in a counterclockwise fashion, where \(a>0\).
For the fundamental theorem to apply, the chosen path must be in the domain of the potential function.
Use the Fundamental Theorem of Line Integrals to compute
\[ \oint _Y \frac {-y}{x^2+y^2}\d x + \frac {x}{x^2+y^2}\d y \]
where \(Y\) is a polygonal path you choose for yourself that contains the point \((0,0)\) in the interior. A square of some sort would probably be easiest.
For the fundamental theorem to apply, the chosen path must be in the domain of the potential function.

At this point, you might suspect that something strange is going on…

Again letting \(\vec {F}(x,y) = \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}}\), compute:
\[ \curl \vec {F} \]
Is \(\curl \vec {F}\) zero?
It is not always zero.
What would you say to Devin and Riley to assure them that mathematics is not “broken?”

1 The take-away

Here we presented you with a field where the (scalar) curl was zero everywhere except at the origin. At the origin the (scalar) curl was undefined; hence, Green’s Theorem does not apply.

What is remarkable is that in this case, where the (scalar) curl is zero except for a point, any path \(C\) around the point where the field is undefined will yield the same value for:

\[ \oint _C \vec {F}\dotp \d \vec {p} \]